1 Answer
1

You have a string of 4 stars and 2 bars (to split it into 3 pieces, one for each variable), there are a total of $\binom{4+2}{2} = \binom{4+2}{4} = \frac{6!}{4!2!} = 15$ such strings, each corresponding to a unique assignment, so there is 15 assignments...

What is meant by "2 bars" ? Is it "k-1" ??
–
Way to infinityFeb 26 '13 at 22:37

@SultanAhmedSagor Yes, in general case, you are looking for ways to get $n=4$ using $k=3$ integers, and the result is $n$ starts, $k-1$ bars, to $\binom{n+k-1}{n} = \binom{n+k-1}{k-1}$.
–
gt6989bFeb 27 '13 at 13:37